## Search found 48 matches

Fri Dec 16, 2016 9:36 pm
Forum: National Math Olympiad (BdMO)
Topic: BdMO National 2016 Secondary 3: Weird angle condition
Replies: 3
Views: 1922

### Re: BdMO National 2016 Secondary 3: Weird angle condition

A synthetic solution : Let $Q$ be a point such that $\triangle PQB$ is equilateral ,$C$ and $Q$ are on the same side of $PB$.Then $Q$ is the center of $\odot BPC$. So, $AQ$ is the perpendicular bisector of $BC$ . As,$\angle APB=150^{\circ}$, so $AP \perp QB \Rightarrow AP$ bisect $\angle QAB$ .So ...
Sat Dec 10, 2016 11:49 am
Forum: Number Theory
Topic: USAJMO 2016
Replies: 1
Views: 1571

### Re: USAJMO 2016

Tue Nov 29, 2016 11:38 pm
Forum: Number Theory
Topic: Infinitely many primes divide $1!+2!+\cdots +n!$
Replies: 1
Views: 1197

### Re: Infinitely many primes divide $1!+2!+\cdots +n!$

Let's prove a more general result: For a prime $P$ ,define $C_p = \{x : P^x \mid S_n ,$ for some $n \in \mathbb N \}$, then $C_P$ is a bounded. Proof : Assume , $C_P$ is not bounded for a prime $P$. Then - Claim 1 : $P^x \mid S_{px}$ ,for any $x \in \mathbb N$ Proof: As $C_P$ is unbounded ,there e...
Tue Nov 22, 2016 4:10 pm
Forum: Geometry
Topic: APMO 2013 P5
Replies: 3
Views: 1813

### Re: APMO 2013 P5

Tue Nov 22, 2016 12:54 am
Forum: Geometry
Topic: A Beauty from Evan Chen
Replies: 1
Views: 1231

Let $AG$ and $MO$ meet $BC$ at $K,S$ respectively .$KM$ meets $PN$ at $R$. As $M$ is the center of $\gamma$,so $MO$ is the perpendicular bisector of $AG$ . $\angle PAM =\angle PGM =90^o \Rightarrow AP\| BC \Rightarrow \angle APM =\angle OSN.$ $ON=\frac {1}{2} AH =AM$ and $\angle ONS =90^o =\angle ... Mon Nov 21, 2016 11:44 pm Forum: Geometry Topic: Two triangles and three collinear points Replies: 1 Views: 1889 ### Re: Two triangles and three collinear points$AD$,$BE$and$CF$concur on the orthocenter$H$of$\triangle DEF$. Let$D_0,E_0,F_0 $be the projection of$D,E,F$on$EF,FD $and$DE$respectively .Let$\alpha$be the circumcircle of$\triangle DEF$. Lemma : The perpendiculars to$BC, CA, AB$through$ R,S,T $respectively intersect at$H$Proo... Sat Oct 15, 2016 7:06 pm Forum: Algebra Topic: Polynomials without real solutions Replies: 1 Views: 2026 ### Re: Polynomials without real solutions My solution: Let$P$and$Q$be the graph of$P(X),Q(X)$.$P,Q$doesn't intersect (as$P(X)$can't be equal to$Q(x)$).Assume$P(P(X))=Q(Q(X))$.Consider the points$A(P(X),P(P(X))),B(P(X),Q(P(X))),C(Q(X),P(Q(X))),D(Q(X),Q(Q(X)))$.$AD \parallel BC \parallel XX^{'}$and$AB \parallel YY^{'} \paralle...
Sun Sep 11, 2016 11:44 pm
Forum: Geometry
Topic: Cyclic intersections of $AH$ and $BO$ gives similar $\Delta$
Replies: 2
Views: 1359